help-10

[misty beacon]

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[hallow socket]

Help with that - if I can get help from a helper that would be lovely

[undone sparrow]

I mean you can just plug in the values into the cylinder area formula

[hallow socket]

My formulas aren’t working

[high lily]

show work

[hallow socket]

I didn’t work it out

I just typed random numbers

[high lily]

then what did you mean by formula's aren't working

[hallow socket]

Like google

It’s just not working for me

[high lily]

wdym by "not working"

do you have the formula in front of you?

[hallow socket]

No so I’m my google isn’t working

It’s hella slow so it’s not loading the formulas - I usually use the internet ones

[high lily]

do you have note? textbook?

[hallow socket]

At school

[undone sparrow]

I think he literally just inputs the heigh and radius into some website to calculate for him

[hallow socket]

Yes I do

I just search up (name) area formula

[undone sparrow]

Can you get the formula and utilize it on an actual calculator

[hallow socket]

No

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[undone sparrow]

Alright then....

My man hates calculators ig

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[unique cobalt]

So I'm trying to solve this question: "Use Euler's phi function to calculate phi(132), then also decide the smallest positive value for m so that m is congruent with 1121^1121 (mod 132)

[unique cobalt]

I got the first part. it's

40

That said, I got stuck on the second part

[slim lake]

So you know that $1121^{40}\equiv 1 mod 132$

[warm shale]

Max..

[unique cobalt]

Aha, so it's for all values?

[slim lake]

Well first actually check for common divisors

Cleary 132=12×11

1121 is not divisible by 2 (clearly)

[unique cobalt]

Actually, since this is a mod function

Can't we rewrite 1121 as 65?

[slim lake]

1+1+2+1=5 so not divisible by 3

[unique cobalt]

It'd be easier to do so, no?

[slim lake]

1-1+2-1=1 so not divisible by 11

So they are coprime so you can use eulers theorem

$a^{\phi(n)}\equiv 1 mod n$

[warm shale]

Max..

[slim lake]

And then just do repeated multiplication

1121^80 is congruent to 1 mod n

So 1121^160 is too ext.

[grand rampart]

Anyone help a brother out

[slim lake]

#❓how-to-get-help

[unique cobalt]

I think I get it, thanks for the help

Gonna give it a try

[slim lake]

👍

[unique cobalt]

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[slim lake]

Yep good idea

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[unique cobalt]

Actually, just to check

[unique cobalt]

So what you're saying is that as long as they have GCD = 1

That means that any number of multiples of the phi of the mod is congruent to 1?

As in the example above, 4*40 is congruent to 1?

[slim lake]

Yes because $a^{phi(n)}\equiv 1 mod n\Rightarrow ($a^{phi(n)})^k\equiv (1)^k mod n$

[warm shale]

Max..
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[unique cobalt]

i SEE

I think'

Does this still work for multiples of a number?

Ie

7 ^9*7?

[slim lake]

For multiples I think you need to use Chinese remainder theorem

But I'm not sure, I'll check later

[unique cobalt]

Thanks

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[left needle]

sorry for bad translation,
A point which isnt an element of the set of points in a complex plane which are given by the following system of inequalities |z+i|>2 and |z-1|<2, where z is a complex number

1. (1,-1) 2) (0,-1) 3) (1,0) 4) (1,1)

i got y>-x
so thenboth 1 and 2 dont belong in the set
but in the solutions its 4)
which im pretty sure isnt because also sqrt(1^2+2^2)>2 and sqrt((1-1)^2+1^2)<2 so it does belong
also if i use the definition of modulo i get that 1) and 2) dont belong, in first case its the first condition and in the second case its the second condition
could someone tell me if its correct that 1) and 2) dont belong in the set

[obtuse pebble]

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[kind hawk]

what is the original language?

[obtuse pebble]

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[left needle]

croatian

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<@&286206848099549185>

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[weak beacon]

hi, how do i prove tan(a+b)/tan(a-b) = [sin a cos a + sin b cos b]/[sin a cos a - sin b cos b]?

[weak beacon]

<@&286206848099549185>

[boreal condor]

write in sin and cos

expand

and simplify

[weak beacon]

on the left side? no?

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[hallow zephyr]

For the logistic regression model using only those predictors selected by KIC in Question 2.5, use the bootstrap procedure (use at least 5, 000 bootstrap replications) to find a confidence interval for the odds of having heart disease for the 65th and 66th patients in the test data. Use the bca option when computing this confidence interval.
Anyone understand how tf this work?

I know how to bootstrap the entire dataset and calculate the confidence interval in R, but how do you bootstrap only one object?

[hallow zephyr]

Also would help a lot as well if you somehow know how to do this in R

ive been working on this for 3 days and am about to explode

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[timid silo]

can someone teach me pls

[timid silo]

im a primary six student

[hexed fern]

uh so you need to find the volume of the small cuboid first

and then the volume of the big cuboid

then divide the volume of the big cuboid by small cuboid

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[blazing jasper]

?

it sppouse to be sin(90 -a) = cos(a) not sin(90 +a) = cos(a) I think

[tranquil arch]

sin(90°+a)=cos(a) is also correct

[blazing jasper]

really?

dose this count to all of this

it matter the minus?

[tranquil arch]

,w sin(90°+a)

[warm shale]

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[blazing jasper]

mm ok

,w cos(90°+a)

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[blazing jasper]

,w tan(90°+a)

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[blazing jasper]

ok

thanks

[blazing jasper]

thank you

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[sage dagger]

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[shadow lava]

Why is this giving horizontal asymptote @ y=3? I thought in order to get horizontal asymptote the denominator power must be larger than than the numerator power?

And is that leading term power only? Or combined numerator power vs denominator power? .. either way I see the numerator and denominator degrees being equal…

[wild swallow]

the horizontal asymptote is at y = 3 precisely because the numerator has 3x^2 and the denominator has x^2

[shadow lava]

I get that they cancel out.. but powers are still equal on top and bottom of the rational function are they not?

[wild swallow]

yes thats why theres an asymptote at y=3

[shadow lava]

It here for 1) it says denominator power must be higher than numerator power to have a horizontal asymptote

[wild swallow]

no its greater than or equal to

[shadow lava]

Oh I should adjust my notes

Ty

I added those myself in light blue but not quite correct., thanks

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[shadow lava]

.reopen

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✅

[shadow lava]

Oh yeah.. is that leading term degree only?

[wild swallow]

thats how it is for rational functions

[shadow lava]

OK thanks I will send screenshot soon just to double check

Like this?

[wild swallow]

yes that seems ok

[shadow lava]

Ty!

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[obtuse pebble]

@surreal moth Has your question been resolved?

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[surreal moth]

.reopen

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✅

[surreal moth]

<@&286206848099549185>

[median carbon]

|x-3|+|x+3|-a=0
|x-3|+|x+3|=a

you look for such a that this equation has 3 solutions

[misty mason]

2x = a?

[surreal moth]

-2x=a , 3=a-3?

[median carbon]

case 1. x-3 < 0 and x+3 < 0
x < 3 and x < -3
case 2. x-3 < 0 x+3 ≥ 0
x < 3 and x ≥ -3
case 3. x-3 > 0 x+3 ≥ 0
x ≥ 3 and x ≥ -3

and you use the definition of absolute value

if there's an a such that it's a solution for all these 3 cases then it's the point at which you have the asymptote

[wild swallow]

vertical and horizontal asymptotes included i guess?

[median carbon]

I think, idk maybe im just dumb, i haven't been solving absolute value problems for a while

[surreal moth]

yes

[wild swallow]

just think about it for a moment and it makes no sense

the function is even

and x=0 cannot be a solution without there being a whole connected interval of solutions

[wild swallow]

and the last asymptote is the horizontal asymptote

essentially what you have is two absolute value functions being added

one centred at 3 and one centred at -3

so in the region between -3 and 3 they "cancel" to become flat

so the only values of a where theres 2 solutions is when a > f(0) = 6

[surreal moth]

oh i see

[median carbon]

Thanks snow, and sorry for misguiding you @surreal moth

[surreal moth]

@median carbon no problem

@wild swallowthank you

[wild swallow]

[surreal moth]

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[gritty carbon]

What's the quickest way to solve this

I've tried subbing in z for (x + iy) and then matching the real component and the imaginary component but that just takes way too long

Is there anything else I can do that's remotely quicker

[kind hawk]

Just use the quadratic formula

[gritty carbon]

lmao

one sec

cheers @kind hawk haha, mind is going crazy

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[vagrant sand]

Hi I need Help

[vagrant sand]

Can Anyone Help me to Solve It?

[sacred root]

Ok which questions actually??

[vagrant sand]

b until i

[sacred root]

Ok

Let me see

Let's take B a)

$4 \sin x\cos x= \sin x$

[warm shale]

Basudev

[sacred root]

Yes

So

You can write it as

[autumn coral]

$4 cosx sinx=sinx$

[warm shale]

Estebson

[autumn coral]

Oh you already did it

[vagrant sand]

Yes

I already Finished The First Question

[sacred root]

Ok

[vagrant sand]

But Other Idk

[sacred root]

So

Let's take

i

6tanx-3cotx=7

Now

You can write tanx=sinx/cosx

And cotx=cosx/sinx

Then take lcm

Yes

$\frac{6\sin(x)^2-3\cos(x)^2}{\sin(x)\cos(x)} = 7$

[warm shale]

Basudev

[vagrant sand]

Ok

[sacred root]

Then take 3 common and write sin(x)² = 1-cos(x)²

[warm shale]

Basudev

[sacred root]

Well

I see now way that leads to a answer

Ye

Instead of that

Try this

$6\tan(x) - \frac{3}{\tan(x)} = 7$

[warm shale]

Basudev

[sacred root]

Let tanx be a

Them

$6a²-3=7a$

Then solve for a

[warm shale]

Basudev

[sacred root]

Solve for a

[vagrant sand]

-1/3,3/2

[sacred root]

Ye

[obtuse pebble]

@vagrant sand Has your question been resolved?

[vagrant sand]

Yes

[obtuse pebble]

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[unique sandal]

Hi guys, can anyone explain the solution for this question. Show that cos54/cos18=2sin18.

[high lily]

which part of the solution do you have be an issue with

[unique sandal]

I don't understand why cos54 became sin(90-54)

[high lily]

complementary identity

[unique sandal]

Ok, right. Thanks:)

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[cold thistle]

,rccw

[tawdry cove]

I didnt know how to solve b

[cold thistle]

A parallelogram is a four-sided shape where the opposing sides are parallel to each other

So you'll want to show that HG and EF are parallel, and that EH and FG are parallel

[tawdry cove]

thats it?

but the question is show that EIGK

[cold thistle]

To show that it's a parallelogram yup

[tawdry cove]

not EFG

[cold thistle]

Oh I didn't see that bit

[tawdry cove]

EFGH*

[tawdry cove]

thats a given

[cold thistle]

So you'd want to construct the sides IG and KE and then prove that they're parallel, along with showing EJ and GK are too

Yeah I thought that was odd since it was given

[tawdry cove]

when i finish i have to write so efgh is a parm (each 2 opp sides are parallel)?

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[ornate smelt]

How do u find the range of sinx-cosx

[high lily]

try harmonic trig identities

to express that as a single trig function

[ornate smelt]

Uhm can u pls explain in detail from the first principles

[high lily]

probably easier to look up harmonic trig identities

[ornate smelt]

Like is it possible by finding the slope/derivative and then finding maxima and minima

[spice citrus]

yes, that's totally possible

[obtuse pebble]

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[cobalt hatch]

Hello, can anyone tell me the area of corn and how they got it?

[high lily]

the diagram screams invalid,
there's an issue with this question

[rotund rune]

the diagram pains my eyes

[cobalt hatch]

I got help from someone and they got this answer

Idk how they got it

[covert crag]

Why?

[high lily]

on the left rectangles, they ** cant** have a width of both
x and 5x+2

[rotund rune]

x = 5x + 2
4x = -2
x = -1/2

[covert crag]

Hmmmm

[rotund rune]

[cobalt hatch]

I need thw answer and i gotta explain it 😔😔

[high lily]

combine what nich and I wrote,

and don't bother with what the question wants from you

[cobalt hatch]

This a project tho 😭😭

[covert crag]

What if they want the answer to be numerical?

[high lily]

x = 5x + 2
4x = -2
x = -1/2

can't have negative length

[rotund rune]

painful to see

[covert crag]

Yea true

[high lily]

unless you want to isolate some laws of physics or something

[cobalt hatch]

Hoq the hell

[daring rock]

🤷 Your teacher made a mistake. The problem just doesn't make sense. All you can do is explain why.

[cobalt hatch]

Aight thanks boys

[rotund rune]

imaginary field

[true berry]

this looks fun lemme try

[true berry]

oh shi

just when i had hope

but ill still try

[cobalt hatch]

Try

[true berry]

wish me luck brotha

[cobalt hatch]

Gl

[true berry]

oh i checked it out

i can see why people said it wouldnt work

[cobalt hatch]

Wsg

[true berry]

so

[cobalt hatch]

Is it because its an irregular shapw

[true berry]

yeah

i did try to work around it

but then id have to use the quadrilateral are a base

you fine with that?

[cobalt hatch]

Yeah

[true berry]

but gotta assume both the squash and pumpkins are quads

so will it even help?

[cobalt hatch]

Idk man it's so confusing 😭

[true berry]

yeah it dont work chief

[cobalt hatch]

Rip

[true berry]

lemme try one last time

o

k

@cobalt hatch its seems solvable now

nah im done

y = 0

😦

[obtuse pebble]

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[wintry zealot]

why do we call
,tex x^{2}
(quad func) 'a polynomial'
,tex \left|x\right|
(abs func) 'not polynomial'
is there any point I missed, both of them has Range > 0 in spite one of it being linear

[wintry zealot]

nvm

[fossil crag]

A polynomial is a finite sum of terms alike a * x^k, with "a" a constant and k a non-negative integer

You will not be able to find a finite sum of those terms such that |x| = this sum

[wintry zealot]

sorry, wdum by infinite , sum ?

[timid silo]

$a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0$

[warm shale]

rbit ✨

[wild swallow]

you can possibly find an infinite sum but polynomials need to be finite

[wild swallow]

also your polynomial is backwards

[timid silo]

does it matter

do you write x²+1 or 1+x²?

[wintry zealot]

i mean what makes abs fun infinite there's just 1 variable at least as in my writings

[wild swallow]

it does when you write power series and then p-adic numbers etc etc

nobody writes their power series from infinity downwards

[drifting swan]

Good morning, I hope you are feeling well. Can you please help me with the following question?
what does the limit of (a(n))^(1/n) with the limit of a(n+1)/a(n) ?

[wild swallow]

alien

#❓how-to-get-help

[timid silo]

but we talking about finite stuff

idk where he got the infinite from

[wild swallow]

gotta consistent ya know

[timid silo]

this is how i learned

[wild swallow]

[wintry zealot]

really confused I though you guys said abs fun has infinte sum

[timid silo]

there is an infinite sum to get it yes, but regular polynomials are finite and can at most approximate it

[wild swallow]

is there an infinite sum to express |x| tho on its full domain?

[timid silo]

not the full domain i think

[wintry zealot]

still doesn't get the idea why abs fun has infinite sum

[timid silo]

maybe its best to just forget about infinite for now

[wild swallow]

you could possibly do something by expanding it wrt a basis of R[[x]]

[timid silo]

we cannot express it like this, so its not a polynomial

$\abs{x} = \sqrt{x²}$

[warm shale]

rbit ✨

[wild swallow]

make R[[x]] a hilbert space and then use orthogonal projection theorem perhaps

[timid silo]

we can express it like this, but thats not a polynomials

[wintry zealot]

sorry if you guys don't mind can you elaborate it in more algebra 2 way, i am having hard time understanding the new terms like mat

[timid silo]

dont give me words beyond analysis I

[wild swallow]

ignore that lmfao

[wintry zealot]

xd

[wild swallow]

basically

polynomials are infinitely differentiable

and

|x| is not differentiable at x=0

so it doesnt work

[timid silo]

thats very algebra lol

[wild swallow]

polynomials are "smooth"

and |x| is not "smooth" at x=0

so it doesnt work

[timid silo]

nice

[wild swallow]

[wintry zealot]

[wintry zealot]

u am kinda getting into it

it could be the reason why it makes v shape

w/c is super sharp

[wild swallow]

its kinda

hard to show it otherwise

like

when you start asking questions of whether or not a function can be expressed as a polynomial

you get into like

how well polynomials can approximate functions

and then you very quickly venture into the realm of like

vector spaces and norms and stuff

[wintry zealot]

yea ofc

btw

[timid silo]

idk, in this case shouldnt be tooo hard to prove it algebraicly that its not possible right?

[vagrant gale]

f(x) = |x| is not differentiable at x=0 because limx->0-≠limx->0+

[timid silo]

because the limits of
|h| / h are different

[wild swallow]

well then thats calculus lol

i thought we wanted algebra

[wintry zealot]

thanks @vagrant gale i kinda have no idea about limits

[wintry zealot]

yea ofc

[vagrant gale]

damn I'm just trying to help

[timid silo]

i was thinking of some prove by contradiction, assume there is a polynomial with degree ≥ 2, and then just find an x large enough so that the value is larger than |x| or something

[wintry zealot]

like this https://www.desmos.com/calculator/zxczgfuzlu

[velvet swan]

isnt |x| = x?

[kind hawk]

|-2|=2

[velvet swan]

yes but like |x| = sqrt(x²) means that |x| = x

the squaring and sqrt of x cancel out

which leaves |x| = x

[kind hawk]

no

that's the point

they don't cancel

$\sqrt{(-2)^2} = \sqrt 4 = 2 \neq -2$

[warm shale]

Denascite

[wintry zealot]

why, how ?

@wild swallow and @timid silo for the help

[kind hawk]

|x| is not equal to x

[obtuse pebble]

@wintry zealot Has your question been resolved?

[obtuse pebble]

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[placid yarrow]

Trying to simplify this (apparently the answer is csc(t), but I’m pretty lost. Sorry if this work is hard to follow lmao

[placid yarrow]

(the circle numbers are just my attempt to show the order of my work)

[final thunder]

You can’t cancel $\frac{1-\sin ^2 t}{\sin t}$ down to $1-\sin{t}$

[warm shale]

Pure

[placid yarrow]

Ooooh ok

[final thunder]

Yeah combine the fractions instead

[placid yarrow]

OOOH

the (Cos^2(t)/Sin(t)) and the (Cos(t)Sin(t)/Cos(t))?

[final thunder]

I meant $$\frac{1-\sin ^2 t}{\sin t} + \sin{t}$$ at the end

[warm shale]

Pure

[final thunder]

This will get you to the answer straight away

$$\frac{1-\sin ^2 t}{\sin t} + \sin{t}= \frac{1-\sin ^2 t}{\sin t} +\frac{\sin ^2 t}{\sin t} $$

[warm shale]

Pure

[final thunder]

=1/sint

=csct

$\blacksquare$

[placid yarrow]

Oooh, just to clarify, you get from the lone sin(t) to sin^2(t)/sin(t) by setting a denominator of 1 and multiplying it by sin(t) to make a common denominator right

[final thunder]

Yes

[warm shale]

Pure

[final thunder]

Whatever

[placid yarrow]

OOH i GOT IT NOW

Thank you!

How do I tell the bot the question is solved without waiting for it to ask

OH WAIT

.close

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[timber dock]

Can someone help me with this?

[rigid lintel]

where are you stuck

[timber dock]

Guess you could say everywhere

[obtuse pebble]

@timber dock Has your question been resolved?

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[ruby oyster]

i am learning suvat equations

[ruby oyster]

it says the average velocity is 0.5(u+v) where u is initial speed and v is final speed

why is this the case

surley a better result for average velocity would be to take velocity readings at intervals

for example is u=10 and v=20 using the first equation average velocity is 15

but if we take velocity readings at equal intervals because acceleration is constant when velocity reads 10,12,14,16.18.20m/s

our avergae velocity is 13.888 or something

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[timid silo]

How i can distribute -7 - ( -n - 6 ) ?

[latent walrus]

3n-7--n--6=3n-7+n+6

[timid silo]

i'm confused

can you draw it instead ?

[latent walrus]

if you subtract a negative it will become a positive, also if you multiply a negative by a negative it becomes a positive. Here we have -(-n-6), you can also think of it as -1(-n-6) so multiply the terms inside by -1 to give (n+6) then add that to 3n-7

[high lily]

consider
$$=3n - 7 - \gray{1}(-n-6)$$
its just unnecessary to write that 1

[warm shale]

ℝamonov

[timid silo]

thank you

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[plucky patio]

Given a quadratic, is there an easy way to find the other quadratic that multiply together to form a certain quartic

[plucky patio]

Like say for example, an easy way to find x^2-2x+2 as the other factor

[gloomy valve]

polynomial division?

[plucky patio]

lol that takes quite long

[gloomy valve]

but in this case its pretty easy to see

(x^2 - 2x + 2)(x^2 + bx + 2) is already pretty obvious as you need an x^4 term and a 4 as constant

but when its not so obvious, polynomial division is the way ;)

[obtuse pebble]

@plucky patio Has your question been resolved?

[plucky patio]

Is there a quick way you can check if you turned a equation into partial fractions correctly?

given that you also have your calculator

[obtuse pebble]

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[wraith surge]

i dont understand how standard form works

so do u just times them together

ah no i figured it out lmao

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[plucky patio]

.reopen

[plucky patio]

?

[obtuse pebble]

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[thorn hamlet]

If I get money 52 times in 10 seconds,
What is the interval I am getting it at (in milliseconds)

[obtuse pebble]

@thorn hamlet Has your question been resolved?

[thorn hamlet]

<@&286206848099549185>

[tardy epoch]

10 seconds = 10 * 1000 milliseconds

[thorn hamlet]

?

[tardy epoch]

what's confusing?

1000 milliseconds = 1 second

[thorn hamlet]

10 x 1000?

[tardy epoch]

• means multiplication, yes

[thorn hamlet]

So what would the interval be

?

[wheat jacinth]

well its 52times/10seconds so its 10seconds/52times

so how many seconds per 1 time

[thorn hamlet]

5.2 per second

But what is the interval to get to that?

[wheat jacinth]

well its 5.2times/second so we get 1second/5.2 times but what i want is (some number of seconds)/1 time

[thorn hamlet]

The interval I’m talking about is like 192 milliseconds

That’s just a random number

[wheat jacinth]

yeah i know but im trying to help you get to the interval

[thorn hamlet]

Ok

[wheat jacinth]

the interval is for example 192milliseconds/1 time right

cause every time takes 192 ms

[thorn hamlet]

Yea

[wheat jacinth]

okay so you have 52times/10seconds which means 10seconds/52 times right

and 10 seconds is how many ms

[thorn hamlet]

10000

[wheat jacinth]

yeah so 10000ms/52times

now i have a fraction that looks like 10000/52 and i want to get it into the form (something)/1

[thorn hamlet]

192.3

[wheat jacinth]

yeah i think thats right

does that makes sense

[thorn hamlet]

Yes

Tysm

[wheat jacinth]

np

[thorn hamlet]

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[obtuse pebble]

@supple raft Has your question been resolved?

[obtuse pebble]

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[runic cargo]

I am having trouble I have followed the steps in my instructions/text book I seem to be getting the wrong answer is there a formula for the calculator

[royal shard]

,w 5*|sec(2pi0.82)|

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ~Martin#3125

[runic cargo]

Perfect thank you!

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[mellow compass]

shouldnt the answer be negative since you have to plug in u

which would lead to 0 - 648/5 pi?

[gleaming saffron]

no

they changed the limits with u=9-y, so u dont need to plug in the equation for u

then since du=-dy, they swapped the limits and cancelled the minus sign

so its 648/5 pi - 0

[mellow compass]

so just to make sure, if you change the limits you dont have to plug in for u = 9-y, except for the limits?

[gleaming saffron]

ye cuz if you really think about it

it gives you the same thing

if you really wanted to, you could leave the limits the same and then plug u=9-y back and itd still give the same answer

try it out

[mellow compass]

yea i just did

nice

ok thanks

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[iron dune]

I am stuck on part B. I am not sure if it has something to do with the tangent secant theorem.

[gloomy valve]

as far as i can tell pythagorean theorem + similarity suffices

By the pythagorean theorem we have
$$DB^2 = DC^2 + CB^2 \quad \text{and} \quad DB^2 = DA^2 - AB^2$$
By similarity you get
$$\frac{AB}{DB} = \frac{DC}{BC}$$

[warm shale]

nvx

[gloomy valve]

You can eliminate DC & AB in the system

[iron dune]

oh

thanks

is this the only way to solve it?

[gloomy valve]

idk

unfortunately, i dont know any secant-tangent circle theorems :(

but its not too bad using pythag theorem & similarity

[iron dune]

What does this mean?

[gloomy valve]

one sec

So let's take the first equation
$$DB^2 = DC^2 + CB^2$$
we have to get an expression only involving $DB$, $DA$ and $CB$. We can use the similarity equation to get $DC = \frac{AB \cdot BC}{DB}$

[warm shale]

nvx

[gloomy valve]

If we plug this into the first equation again:
$$DB^2 = \frac{AB^2 \cdot CB^2}{DB^2} + CB^2$$

[warm shale]

nvx

[gloomy valve]

Now you can eliminate the $AB^2$ term using the second equation $DB^2 = DA^2 - AB^2$

[warm shale]

nvx

[iron dune]

how does it eliminate it?

[gloomy valve]

$DB^2 = DA^2 - AB^2 \iff AB^2 = DA^2 - DB^2$

[warm shale]

nvx

[gloomy valve]

Plug this here for AB^2

$DB^2 = \frac{(DA^2 - DB^2) \cdot CB^2}{DB^2} + CB^2$

[warm shale]

nvx

[gloomy valve]

If you rearrange this equation to have DB^2 one on side, you get the result

[iron dune]

ty I understand now

so it ends up with db^4=DA^2*CB^2

and then square root it

[gloomy valve]

Yes!

[iron dune]

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[ocean bane]

im doing question b

i tried squaring the numerator and the denominator but that kinda trumped me

[plain stag]

I wouldn't square both, but multiply by the conjugate of the denominator

[ocean bane]

oooohh rationalizing gotcha

[plain stag]

Yep

Kill the sqrts and hopefully the denominator will be nicer

The numerator will be uglier but we'll cross that bridge later

[ocean bane]

yup! one moment

should i rationalize the numerator as well?

[plain stag]

The numerator is fine

[ocean bane]

kk

[plain stag]

I think that x-1 is quite cleverly chosen

(Assuming my mental math is right, which isn't guaranteed)

[ocean bane]

yup! when i factored the denominator i was able to cross out the x-1 from both

[plain stag]

Ok good, that's what I hoped for

[ocean bane]

hmmm

my ans is sqrt(3) + sqrt(3) / 3

ive just checked and that is indeed the answer. thank you!

[plain stag]

[ocean bane]

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[obtuse pebble]

@lime delta Has your question been resolved?

[obtuse pebble]

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[unborn canyon]

I have no clue how to do this like 0 clue and I need a nudge in the right direction

[royal shard]

look at the definition of the derivative

the limit definition

then use the property

@unborn canyon

are you familiar with this definition?

$f'(x)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$

[warm shale]

~Martin

[royal shard]

here we can split up the f(a+h) into f(a)+f(h)+2ah

you should be able to do the rest

[obtuse pebble]

@unborn canyon Has your question been resolved?

[unborn canyon]

so just replace the first f(a + h) into f(a) + f(h) + 2ah?

[royal shard]

yes

[unborn canyon]

alrght ill be back in a sec once itry

thanks for the help

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[unborn canyon]

.reopen

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✅

[unborn canyon]

im completely lost 😭

[royal shard]

where are you lost?

[unborn canyon]

like from the beginning....

[royal shard]

this is the definition

you get that?

[unborn canyon]

yes

[royal shard]

this is given in the problem

applying it, we arrive here

notice that f(a)-f(a)=0

so it cancels

so we are left with this

now split the fraction apart

now use the second information given

about the limit

[unborn canyon]

OH

THANKS SO MUCH....

[royal shard]

you're welcome^^

[obtuse pebble]

@unborn canyon Has your question been resolved?

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[rapid marsh]

need to solve this using formula

but I am having trouble

[earnest elk]

So, 144*pi = pi * r^2

Yes?

Find the radius and using the radius you can find the diameter

[rapid marsh]

144(3.14)=3.14r^2

[earnest elk]

You can just divide both sides by pi

Don’t need to use approx values

It will cancel out nicely

[rapid marsh]

24

[earnest elk]

Yes that is correct

[rapid marsh]

I've been forgetting a lot of basic stuff recently 💀

I appreciate the help

[earnest elk]

No problem

Have a nice day

[rapid marsh]

you as well

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[molten cosmos]

Is R^omega injective?

I'm not sure if I'm understanding the right shift correctly

does it just shift every element up 1 index and then enter the value 0 at the first position?

Also this isnt surjective right because we cant map to any sequence that doesnt start with a 0

[obtuse pebble]

@molten cosmos Has your question been resolved?

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[full heart]

How do I attempt to do this problem??how can I simpify it I just get to (6w^5-192)/w-2 and it just feels impossible to get anywhere else.

[full heart]

There's no way to factor out a w-2 as far as I can see :/

[lethal sand]

note that 192 is just 6*2^5

now just factor out 6(w-2)

you can use long division to find the other factor

or if you are not in for it, you could use the geometric sum formula to derive the factorisation of x^n-1

but honestly, it's just power of 5, long division wouldnt be too bad

[violet sentinel]

Another way is that the diff of powers has a pattern. as well. Like waler said, you can factor out a 6(w-2) at some point. To justify it, consider

$f'(x) = 6x^5 + 1\$
$f'(w) = 6w^5 + 1\$
$f'(2) = 6(2^5) + 1\$
$f'(w) - f'(2) = 6w^5 + 1 - (6(2^5) + 1)\$
$6(w^5 - 32)\$
$6(w - 2)(...)$

[warm shale]

MellowDramaLlama

[violet sentinel]

and so the (...) will be following the pattern we see here

namely, $6(w - 2)(w^4 +w^32 + w^22^2 + w2^3 + 2^4)$

[warm shale]

MellowDramaLlama

[violet sentinel]

now you can cancel out the numerator and denominator, plug in w = 2 and Bob's your uncle

[full heart]

thanks!

[gray fjord]

Hey I’m having a little trouble with this problem, could I maybe have some help in this please so I can understand it better

[obtuse pebble]

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[sage dagger]

Could I get some help on these questions? I just want to clarify the number 5 because I’m a bit confused

[sage dagger]

Say for example c

Is it

x(-x) / x(-x)^2 + 1

[late stump]

no, essentially you replace x with -x

f(x) means that you input a value x

so f(1) means you input the value 1

similarly, f(-x) means you input the value -x

[obtuse pebble]

@sage dagger Has your question been resolved?

[timid silo]

do someone on this channel knows how to solve this?

[obtuse pebble]

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[sage dagger]

So it would be like -x / -x^2 + 1

[obtuse pebble]

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[ornate sigil]

5a + 14b + 9c + 4?

[ornate sigil]

What is the constant in the expression: 5a + 14b + 9c + 4?​

T_T

[dapper bloom]

Can you tell is the exact wording of the question?

[ornate sigil]

hi

[dapper bloom]

Like are you looking for the constant term in the expression?

[ornate sigil]

i'm not good at math...
wait imma ss

oh wait wrong question

[dapper bloom]

Is this a test?

[ornate sigil]

yes

[dapper bloom]

Ah, people cannot help with that. It is against the rules.

[ornate sigil]

passing score is 50

[ornate sigil]

why

i dont understandd

[dapper bloom]

Because it's cheating

[ornate sigil]

oh okay...

well teach mee algebra
i dont understand algebra that much in class

[dapper bloom]

You can ask general algebra questions and stuff but we can't help you on tests and junk.

See #rules

[ornate sigil]

oh okay

oh i cant...

[dapper bloom]

Ideally you would ask all the questions to learn and understand whatever algebra you need before the test

[ornate sigil]

no test

oh okay

[dapper bloom]

Then there's no cheating concerns to be had

[ornate sigil]

😔 😔

oky

[true berry]

i-

this is funny

what was the question

[dapper bloom]

Just some algebra

[true berry]

oh wait

i read the convo

good luck byerno

[obtuse pebble]

@ornate sigil Has your question been resolved?

[ornate sigil]

NO 😡

im going to cook you

[tardy epoch]

can you not cheat on tests, thanks

.close

[obtuse pebble]

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[earnest hearth]

hey

[obtuse pebble]

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[timid silo]

Hi

[timid silo]

I need help with this equation

[true berry]

is it a quadrilateral or rectangles?

or a square

[timid silo]

A parallelogram

ABCD

Hello

Is there anyone who can help me

[true berry]

oh sorry lol

you didnt ping me

m is the intersection right?

[timid silo]

It doesn’t say anything. It just says m is is in the parallelogram

It says “ having a parallelogram ABCD show that {the equation}”

Show that MA + MB - MC - MD = - 2 AD

[true berry]

ok so

[timid silo]

Ok so

Is this correct

I Said that because AD = BC, AD + BC = 2 AD

[true berry]

nooo dude

[timid silo]

No?

[true berry]

im trying to solve this

m is mostly the intersection

[obtuse pebble]

@timid silo Has your question been resolved?

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[urban totem]

Something's wrong, I can feel it...

[shy vigil]

Hmm, so if its x minutes

then first 5 minutes are 5 and rest is x-5

[daring rock]

Wen you say something's wrong, do you mean you think none of theses answers are right?

[urban totem]

Yep

[daring rock]

Yep

[shy vigil]

that is, $f(x)=50+(3+5)(x-5)=50+3x-40=3x+10$

[warm shale]

n11

[shy vigil]

one sec

[daring rock]

I suspect they want you to choose the last answer, but it's not right

[urban totem]

Wait... If its 3x+10 what about the first 5 minutes?

[shy vigil]

actually I made a mistake here, I think

[urban totem]

So they deliberately want me to fail this?

[shy vigil]

it should be (3+10) and not (3+5)

[daring rock]

No, whoever wrote the question made a mistake

[shy vigil]

yeah

[daring rock]

3x+10 would be correct if it said there was a flat fee of $10 before anything else, plus $3 per minute

It says $10 for the first five minutes, then $3 per minute after that. You'd have to make a piecewise function for that

[shy vigil]

At first I thought it says: $10 for the first 5 minutes, then $10 PLUS $3 additionally, for the subsequent minutes

[daring rock]

That's what I thought for a second too, but that still doesn't match any of the answers

[shy vigil]

Yeah

[daring rock]

The only option with a slope of 3 is the last one, and it's not correct

So 🤷

[urban totem]

Thanks @shy vigil @daring rock

It's fine, i think i did good enough on the quiz... It was about rational functions

[daring rock]

Rational functions? That doesn't seem to match this problem

But

I'd ask your teacher about this one. Point out that none of the answers are right so you don't lose points you don't deserve to

[urban totem]

Ikr
This is why many students fail

.close

[obtuse pebble]

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[kindred mango]

does anyone know bout the ration zero theorem?
and if u guys do know
what's the answer for this??????
x³+2x-8=0
ive been trying for like 50 mins now

[shy vigil]

let me see whats the rational zero theorem

ah okay

so you go hit and try and

[high lily]

rat root/zero theorem only helps find rational roots if they exist

x³+2x-8 = 0
doesn't have such roots, so the theorem fails in finding one.

[kindred mango]

so there's no answer for that?

[shy vigil]

for example take x=a, and put it in the polynomial. If the result is zero, then it means x-a is a factor of the polynomial

[high lily]

ther'll be solutions,

there just won't be rational ones

[kindred mango]

thank u🐸 🙏

[high lily]

its in the form of a depressed cubic so i suppose you can do stuff with cardanos

and cubic formula

[random depot]

Depressed

[true berry]

ahhh wrong sever to help you with that

wait wrong member

[obtuse pebble]

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[timid silo]

Can someone write out the middle steps

[gleaming saffron]

you can cancel a term already

two terms infaxt

[timid silo]

How

[gleaming saffron]

theyre equal to each other

[timid silo]

Yeah so what

2=2

Doesnt mean 2 and 2 cancels out

[gleaming saffron]

jesus

[timid silo]

Cute pfp btw

Ikik

[high lily]

multiply or divide both sides by same amount

[gleaming saffron]

thanks bro

[timid silo]

ok

You see what im doing there?

You see that?

thats what im stuck with

[high lily]

you're not multiplying properly

[timid silo]

Yeh

I forgot how to do it properly

Help me

[high lily]

multiply or divide both sides by same amount
explicitly write what you intend to do to each side if needed

note that if you were to actually multiply both sides by $\frac{m_2}{r}$, the right side turns into
$$\frac{m_2V}{r} \cdot \frac{m_2}{r} = \frac{{m_2}^2V}{r^2}$$
and not V

[warm shale]

ℝamonov

[high lily]

recall what you'd do in a simpler example

how would you isolate /solve for $x$ in
$$\frac x2 = 3p$$

[warm shale]

ℝamonov

[timid silo]

Yeah

3px2

6p

[high lily]

don't use x for multiplication,
but yeh, here you'd multiply both sides by 2 to get x

[timid silo]

Yeh

[high lily]

and what if you had
$$3y = 2k$$
how would you solve for $y$ here?

[warm shale]

ℝamonov

[timid silo]

2k/3

[high lily]

yes

[timid silo]

$$yes$$

[warm shale]

zesub

[high lily]

now apply those principles to the equation you have

[timid silo]

Okay

[high lily]

do two separate steps if needed

[timid silo]

$$thank you>100000..$$

[warm shale]

zesub

[timid silo]

.close

[obtuse pebble]

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[vital phoenix]

Hi, please explain to me how you get the rhs from lhs, like the steps taken, Thank you. Formula is 1+w+w^2+ .... + w^(q-p) = (1-w^(q-p+1))/(1-w)

[kind hawk]

multiply both sides by 1-w

its just a famous formula essentially

check out geometric sums/series

[vital phoenix]

does it have a name?

thank you btw

[kind hawk]

.

[vital phoenix]

okey will check it out, ty

.close

[obtuse pebble]

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[modern cradle]

How do I reflect a function over a line?

For example, if I had f(x) how would I reflect that on the line y=3, or x=-2 and so on

[obtuse pebble]

@modern cradle Has your question been resolved?

[ornate smelt]

U do know how to reflect it on the axis ryt

Use graph transformations

[obtuse pebble]

@modern cradle Has your question been resolved?

[obtuse pebble]

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[sage dagger]

Is x + y^2 = 1 a function

[valid jetty]

hi

[sage dagger]

hello

do u know